📝 题目
例 2 设 $f\left( {x,y}\right)$ 是二阶连续可微函数. 试计算
$$ I = {\iint }_{\left\lbrack {a,\beta }\right\rbrack \times \left\lbrack {\gamma ,\delta }\right\rbrack }\frac{{\partial }^{2}f}{\partial x\partial y}\left( {x,y}\right) \mathrm{d}\left( {x,y}\right) . $$
💡 答案与解析
解 化为累次积分计算, 我们得到
$$ I = {\int }_{a}^{\beta }\mathrm{d}x{\int }_{\gamma }^{\delta }\frac{{\partial }^{2}f}{\partial x\partial y}\left( {x,y}\right) \mathrm{d}y $$
$$ = {\int }_{\alpha }^{\beta }\left( {\frac{\partial f}{\partial x}\left( {x,\delta }\right) - \frac{\partial f}{\partial x}\left( {x,\gamma }\right) }\right) \mathrm{d}x $$
$$ = f\left( {\beta ,\delta }\right) - f\left( {\alpha ,\delta }\right) - f\left( {\beta ,\gamma }\right) + f\left( {\alpha ,\gamma }\right) . $$